GCD Expectation

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Time Limit: 4 Seconds     Memory Limit:
262144 KB

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Edward has a set of n integers {a₁,a₂,...,a_n}. He randomly picks a nonempty subset {x₁,x₂,…,x_m}
(each nonempty subset has equal probability to be picked), and would like to know the expectation of [gcd(x₁,x₂,…,x_m)]^k.

Note that gcd(x₁,x₂,…,x_m) is the greatest common divisor of {x₁,x₂,…,x_m}.

Input

There are multiple test cases. The first line of input contains an integerT indicating the number of test cases. For each test case:

The first line contains two integers n,k (1 ≤
n, k ≤ 10⁶). The second line containsn integers
a₁, a₂,…,a_n (1 ≤a_i ≤ 10⁶).

The sum of values max{a_i} for all the test cases does not exceed 2000000.

Output

For each case, if the expectation is E, output a single integer denotesE · (2ⁿ - 1) modulo 998244353.

Sample Input

1

5 1

1 2 3 4 5

Sample Output

42

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Author: LIN, Xi

Source: The 15th Zhejiang University Programming Contest

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?

problemId=5480

题目大意：给一个集合。{xi}为它的一个非空子集。设E为[gcd(x₁,x₂,…,x_m)]^k  
的期望，求E*(2^n - 1) mod 998244353

题目分析：首先一个有n个元素的集合的非空子集个数为2^n - 1，所以E的分母就是2^n - 1了。因此我们要求的仅仅是E的分子,

设F(x)为gcd(xi) = x的个数，那么ans = (1^k) * F(1) + (2^k) * F(2) + ... + (ma^k) * F(ma)

以下的问题就是怎样高速的计算F(x)了。对于一个集合，先计算出x的倍数的个数，nlogn就可以。然后就是基础的容斥。如果如今要求gcd为1的，那就减去gcd为2的，gcd为3的，注意到6同一时候是2和3的倍数，也就是6的倍数被减了两次，所以要加上gcd为6的，前面的系数刚好是数字相应的莫比乌斯函数，看到这题非常多用dp来容斥的，事实上本质和莫比乌斯函数一样，可是莫比乌斯函数写起来真的非常easy。2333333

#include <cstdio>

#include <cstring>

#include <algorithm>

#define ll long long

using namespace std;

int const MOD = 998244353;

int const MAX = 1e6 + 5;

ll two[MAX];

int p[MAX], mob[MAX], num[MAX], cnt[MAX];

bool noprime[MAX];

int n, k, ma, pnum;

void Mobius()

{

    pnum = 0;

    mob[1] = 1;

    for(int i = 2; i < MAX; i++)

    {

        if(!noprime[i])

        {

            p[pnum ++] = i;

            mob[i] = -1;

        }

        for(int j = 0; j < pnum && i * p[j] < MAX; j++)

        {

            noprime[i * p[j]] = true;

            if(i % p[j] == 0)

            {

                mob[i * p[j]] = 0;

                break;

            }

            mob[i * p[j]] = -mob[i];

        }

    }

}

ll qpow(ll x, ll n)

{

    ll res = 1;

    while(n != 0)

    {

        if(n & 1)

            res = (res * x) % MOD;

        x = (x * x) % MOD;

        n >>= 1;

    }

    return res;

}

void pre()

{

    Mobius();

    two[0] = 1;

    for(int i = 1; i < MAX; i++)

        two[i] = two[i - 1] * 2ll % MOD;

}

int main()

{

    pre();

    int T;

    scanf("%d", &T);

    while(T --)

    {

        memset(num, 0, sizeof(num));

        memset(cnt, 0, sizeof(cnt));

        ma = 0;

        int tmp;

        scanf("%d %d", &n, &k);

        for(int i = 0; i < n; i++)

        {

            scanf("%d", &tmp);

            cnt[tmp] ++;

            ma = max(ma, tmp);

        }

        for(int i = 1; i <= ma; i++)

            for(int j = i; j <= ma; j += i)

                num[i] += cnt[j];       //求i的倍数的个数

        ll ans = 0;

        for(int i = 1; i <= ma; i++)    //枚举gcd

        {

            ll sum = 0;

            for(int j = i; j <= ma; j += i)  //容斥

                sum = (MOD + sum % MOD + mob[j / i] * (two[num[j]] - 1) % MOD) % MOD;

            ans = (MOD + ans % MOD + (sum * qpow(i, k)) % MOD) % MOD;

        }

        printf("%lld\n", ans);

    }

}